Let $M$ and $N$ be the reflections of $H$ over $H_B$ and $H_A$ respectively. Using Orthocentre-reflection and PoP, $A'H_A\cdot HH_A=AH_A\cdot H_AN=QH_A\cdot H_AP\implies A'\in (PQH)$ $B'H_B\cdot HH_B=BH_B\cdot H_BM=QH_B\cdot H_BP\implies B'\in (PQH)$ Therefore, $A', B', P, Q$ are concyclic.

Let $H$ denote the orthocenter of $\triangle ABC$. $H',H''$ denote the reflection of $H$ over $BC,AC$ respectively. It is well known that $H',H''\in\odot(ABC)$. Now $B'$ being the reflection of $B$ over $H_B$, we have $$H_BB'\cdot H_BH=H_BB\cdot H_BH''=\operatorname{Pow}_{\odot ABC}(H_B)=H_BP\cdot H_BQ\implies B'PHQ\text{ is cyclic.}$$ Also, $A'$ being the reflection of $A$ over $H_A$, we have $$H_AA'\cdot H_AH=H_AA\cdot H_AH'=\operatorname{Pow}_{\odot ABC}(H_A)=H_AP\cdot H_AQ\implies A'QHP\text{ is cyclic.}$$ So from $A'QHP$ and $B'QHP$ cyclic, we get $A'B'QHP$ is cyclic and we are done.

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